If one were to compare these definitions with their Euclidean counterparts (for example in [1]}), one would see immediate differences. Instead of requiring \(fp\)  to be absolutely continuous, we are able to instead restrict \(f^{\prime } \in L^1(dx)\)  so that \(fp\)  is absolutely continuous. In other words, we need not demand absolute continuity in \(fp\)  since it is a consequence of Definition REF . Stating that \(f^{\prime }\)  is differentiable everywhere allows us to write down, explicitly, the Stein operator as a differential operator in terms of \(f\)  — see below. The key difference is that for the definition of the Stein class on \(\mathbb {R}\) , the third condition \( \int _{\mathbb {R}} (fp)^{\prime } dx = 0 \)  is required (cf. [1]}) for \(p \in L^1(\mathbb {R},dx)\) . In the case of the circle, if \(\mathrm {supp}(P)=\mathbb {S}^1\) , this condition is automatically satisfied: if we are to identify \(\mathbb {S}^1\)  with the interval \([-\pi ,\pi )\)  alongside the equivalence relation \(-\pi \sim \pi \) , then \(f(-\pi )p(-\pi )=f(\pi )p(\pi )\)  and so
\( \int _{\mathbb {S}^1}(fp)^{\prime }dx = (fp)\bigg |_{-\pi }^\pi =f(\pi )p(\pi )-f(-\pi )p(-\pi )=0.\) 
